Answer:
the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.
Explanation:
Given;
length of the wire, = 34.9 m
length of solenoid, L = 0.24 m
radius of the solenoid, r = 0.051 m
current in the solenoid, I = 11.0 A
The number of turns of the wire is calculated as follow;
[tex]N = \frac{34.9}{2\pi \times 0.051} = 109 \ turns[/tex]
The strength of the magnetic field inside the solenoid is calculated as follows;
[tex]B = \mu_0 (\frac{N}{L} )I\\\\B = 4\pi \times 10^{-7} \times (\frac{109}{0.24} )\times 11.0 \\\\B = 6.278 \times 10^{-3} \ T[/tex]
Therefore, the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.
